2.8 Simultaneous equations

We make \(y\) the subject of each equation:

\begin{align*} y + x &= 5 \\ y &= 5 - x \end{align*} \begin{align*} y - x^{2} + 3x - 5 &= 0 \\ y &= x^{2} - 3x + 5 \end{align*}

Next equate the two equations and solve for \(x\):

\begin{align*} 5 - x &= x^{2} - 3x + 5 \\ 0 &= x^{2} - 2x \\ 0 &= x(x - 2) \\ \text{Therefore } x = 0 &\text{ or } x = 2 \end{align*}

Now we substitute the values for \(x\) back into the first equation to calculate the corresponding \(y\)-values

If \(x = 0\): \begin{align*} y &= 5 - 0 \\ \therefore y &= 5 \end{align*} This gives the point \((0;5)\).

If \(x = 2\): \begin{align*} y &= 5 - 2 \\ &= 3 \end{align*} This gives the point \((2;3)\).

The solution is \(x = 0 \text{ and } y = 5\) or \(x = 2 \text{ and } y = 3\). These are the coordinate pairs for the points of intersection.

We make \(y\) the subject of each equation:

\[y = 6 - 5x + x^2\] \begin{align*} y - x + 1 &= 0 \\ y &= x - 1 \end{align*}

Next equate the two equations and solve for \(x\):

\begin{align*} 6 - 5x + x^2 &= x - 1 \\ x^{2} - 6x + 7 &= 0 \\ x &= \dfrac{-(-6) \pm \sqrt{(-6)^{2} - 4(1)(7)}}{2(1)} \\ x &= \dfrac{6 \pm \sqrt{36 - 28}}{2} \\ x &= \dfrac{6 \pm \sqrt{8}}{2} \\ x &= \dfrac{6 \pm 2\sqrt{2}}{2} \\ x &= 3 \pm \sqrt{2} \end{align*}

Now we substitute the values for \(x\) back into the second equation to calculate the corresponding \(y\)-values

If \(x = 3 - \sqrt{2}\): \begin{align*} y &= 3 - \sqrt{2} - 1 \\ \therefore y &= 2 - \sqrt{2} \end{align*}

If \(x = 3 + \sqrt{2}\): \begin{align*} y &= 3 + \sqrt{2} - 1 \\ &= 2 + \sqrt{2} \end{align*}

The solution is \(x = 3 \pm \sqrt{2} \text{ and } y = 2 \pm \sqrt{2}\)

We make \(y\) the subject of each equation:

\[y = \dfrac{2x+2}{4}\] \begin{align*} y - 2{x}^{2} + 3x + 5 &= 0 \\ y &= 2x^{2} - 3x - 5 \end{align*}

Next equate the two equations and solve for \(x\):

\begin{align*} \dfrac{2x+2}{4} &= 2x^{2} - 3x - 5 \\ 2x + 2 &= 8x^{2} - 12x - 20 \\ 0 &= 8x^{2} - 14x - 22 \\ 0 &= 4x^{2} - 7x - 11 \\ 0 &= (4x - 11)(x + 1) \\ \text{Therefore } x = -1 &\text{ or } x = \frac{11}{4} \end{align*}

Now we substitute the values for \(x\) back into the first equation to calculate the corresponding \(y\)-values

If \(x = -1\): \begin{align*} y &= \dfrac{2(-1) + 2}{4} \\ y &= 0 \end{align*} This gives the point \((-1;0)\).

If \(x = \frac{1}{4}\): \begin{align*} y &= \dfrac{2(\frac{1}{4}) + 2}{4} \\ y &= \dfrac{\frac{5}{2}}{4} \\ &= \frac{5}{8} \end{align*} This gives the point \((\frac{1}{4};\frac{5}{8})\).

The solution is \(x = -1 \text{ and } y = 0\) or \(x = \frac{1}{4} \text{ and } y = \frac{5}{8}\). These are the coordinate pairs for the points of intersection.

We make \(a\) the subject of each equation:

\[a = 2b + 3\] \begin{align*} a - 3{b}^{2} + 4 &= 0 \\ a &= 3b^{2} - 4 \end{align*}

Next equate the two equations and solve for \(b\):

\begin{align*} 2b + 3 &= 3b^{2} - 4 \\ 0 &= 3b^{2} - 2b - 7 \\ 0 &= \dfrac{-(-2) \pm \sqrt{(-2)^{2} - 4(3)(-7)}}{2(3)} \\ b &= \dfrac{2 \pm \sqrt{4 + 84}}{6} \\ b &= \dfrac{2 \pm \sqrt{88}}{6} \end{align*}

Now we substitute the values for \(b\) back into the first equation to calculate the corresponding \(a\)-values

If \(b = \dfrac{2 + \sqrt{88}}{6}\): \begin{align*} a &= 2\left(\dfrac{2 + \sqrt{88}}{6}\right) + 3 \\ &= \dfrac{2 + \sqrt{88}}{3} + 3 \\ &= \dfrac{11 + \sqrt{88}}{3} \end{align*}

If \(b = \dfrac{2 - \sqrt{88}}{6}\): \begin{align*} a &= 2\left(\dfrac{2 - \sqrt{88}}{6}\right) + 3 \\ &= \dfrac{2 - \sqrt{88}}{3} + 3 \\ &= \dfrac{11 - \sqrt{88}}{3} \end{align*}

The solution is \(b = \dfrac{2 \pm \sqrt{88}}{6} \text{ and } a = \dfrac{11 \pm \sqrt{88}}{6}\)

We make \(y\) the subject of each equation:

\begin{align*} x^{2} + y + 2 &= 3x \\ y &= -x^{2} + 3x - 2 \end{align*} \[y = 4x - 8\]

Next equate the two equations and solve for \(x\):

\begin{align*} 4x - 8 &= -x^{2} + 3x - 2 \\ x^{2} + x - 6 &= 0 \\ (x + 3)(x - 2) & = 0 \\ \text{Therefore } x = -3 &\text{ or } x = 2 \end{align*}

Now we substitute the values for \(x\) back into the second equation to calculate the corresponding \(y\)-values

If \(x = -3\): \begin{align*} y &= 4(-3) - 8 \\ y &= -20 \end{align*} This gives the point \((-3;-20)\).

If \(x = 2\): \begin{align*} y &= 4(2) - 8 \\ &= 0 \end{align*} This gives the point \((2;0)\).

The solution is \(x = -3 \text{ and } y = -20\) or \(x = 2 \text{ and } y = 0\). These are the coordinate pairs for the points of intersection.

We make \(y\) the subject of each equation:

\begin{align*} 2y + x^2 + 25 &= 7x \\ 2y + x^2 + 25 &= 7x - x^{2} + 25 \\ y &= \frac{-x^{2} + 7x + 25}{2} \end{align*} \begin{align*} 2y & = x - 32 \\ y & = \frac{x - 32}{2} \end{align*}

Next equate the two equations and solve for \(x\):

\begin{align*} \frac{x - 32}{2} &= \frac{-x^{2} + 7x + 25}{2} \\ x - 32 &= -x^{2} + 7x + 25 \\ x^{2} - 6x - 57 & = 0 \\ x & = \dfrac{-(-6) \pm \sqrt{(-6)^{2} - 4(1)(-57)}}{2(1)} \\ & = \dfrac{6 \pm \sqrt{36 + 228}}{2} \\ & = \dfrac{6 \pm \sqrt{264}}{2} \end{align*}

Now we substitute the values for \(x\) back into the second equation to calculate the corresponding \(y\)-values

If \(x = \dfrac{6 + \sqrt{264}}{2}\): \begin{align*} 2y &= \dfrac{6 + \sqrt{264}}{2} - 32 \\ y &= \dfrac{70 + \sqrt{264}}{4} \\ &= \dfrac{70 + \sqrt{264}}{4} \end{align*}

If \(x = \dfrac{6 - \sqrt{264}}{2}\): \begin{align*} 2y &= \dfrac{6 - \sqrt{264}}{2} - 32 \\ y &= \dfrac{70 - \sqrt{264}}{4} \\ &= \dfrac{70 - \sqrt{264}}{4} \end{align*}

The solution is \(x = \dfrac{6 \pm \sqrt{264}}{2} \text{ and } y = \dfrac{70 \pm \sqrt{264}}{2}\)

Make \(y\) the subject of both equations

For the first equation we have:

\begin{align*} x^{2} - 1 - y &= 0 \\ y &= x^2 - 1 \end{align*}

and for the second equation:

\begin{align*} y + x - 5 &= 0 \\ y &= -x + 5 \end{align*}

Draw the straight line graph and parabola on the same system of axes:

From the diagram we see that the graphs intersect at \((-3;8)\) and \((2;3)\).

We can solve algebraically to check. Doing this we find the same solution.

The solutions to the system of simultaneous equations are \((-3;8)\) and \((2;3)\).

Make \(y\) the subject of both equations

For the first equation we have:

\begin{align*} x + y - 10 &= 0 \\ y &= -x + 10 \end{align*}

and for the second equation:

\begin{align*} x^{2} - 2 - y &= 0 \\ y &= x^{2} - 2 \end{align*}

Draw the straight line graph and parabola on the same system of axes:

From the diagram we see that the graphs intersect at \((-4;14)\) and \((3;7)\).

We can solve algebraically to check. Doing this we find the same solution.

The solutions to the system of simultaneous equations are \((-4;14)\) and \((3;7)\).

Make \(y\) the subject of both equations

For the first equation we have:

\begin{align*} xy &= 12 \\ y &= \frac{12}{x} \end{align*}

and for the second equation:

\begin{align*} 7 &= x + y \\ y &= 7 - x \end{align*}

Draw the two graphs on the same system of axes:

We also note that this sytem of equations has the followig restrictions: \(x \ne 0\) and \(y \ne 0\)

From the diagram we see that the graphs intersect at \((3;4)\) and \((4;3)\).

We can solve algebraically to check. Doing this we find the same solution.

The solutions to the system of simultaneous equations are \((3;4)\) and \((4;3)\).

Make \(y\) the subject of both equations

For the first equation we have:

\begin{align*} 6 - 4x - y &= 0 \\ y &= -4x + 6 \end{align*}

and for the second equation:

\begin{align*} 12 - 2x^{2} - y &= 0 \\ y &= -2x^{2} + 12 \end{align*}

Draw the two graphs on the same system of axes:

From the diagram we see that the graphs intersect at approximately \((4;-10)\) and \((-2;14)\).

We can solve algebraically to check (and to get a more accurate answer). Doing this gives:

\begin{align*} -4x + 6 &= -2x^{2} + 12 \\ 2x - 3 &= x^{2} - 6 \\ x^{2} - 2x - 9 & = 0 \\ x & = \dfrac{-(-2) \pm \sqrt{(-2)^{2} - 4(1)(-9)}}{2(1)} \\ x & = \dfrac{2 \pm \sqrt{(4 + 36}}{2} \\ x & = \dfrac{2 \pm \sqrt{40}}{2} \end{align*}

Solving for \(y\) gives:

\begin{align*} y & = -4\left(\dfrac{2 \pm \sqrt{40}}{2} \right) + 6 \\ &= -2(2 \pm \sqrt{40}) + 6 \\ & = -4 \pm 2\sqrt{40} + 6 \\ & = 2 \pm 2\sqrt{40} \end{align*}